Is this the double limit?

Calculus Level 2

f ( n , θ ) = ∏ r = 1 n [ 1 − tan ⁡ 2 ( θ 2 r ) ] \large \displaystyle f(n, \theta) = \prod_{r=1}^n \left [1 - \tan^2 \left ( \frac \theta {2^r} \right ) \right]

Suppose we have a function of f ( n , θ ) f(n,\theta) as above, find the value of lim ⁡ θ → 0 lim ⁡ n → ∞ f ( n , θ ) \displaystyle \lim_{\theta\to 0}\lim_{n\to \infty} f(n, \theta) .


The answer is 1.

Rudresh Tomar by Rudresh Tomar , 25, India

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1 solution

Brian Lie
Mar 26, 2018

L = lim ⁡ θ → 0 lim ⁡ n → ∞ ∏ r = 1 n [ 1 − tan ⁡ 2 ( θ 2 r ) ] = lim ⁡ θ → 0 lim ⁡ n → ∞ ∏ r = 1 n cos ⁡ 2 ( θ 2 r ) − sin ⁡ 2 ( θ 2 r ) cos ⁡ 2 ( θ 2 r ) = lim ⁡ θ → 0 lim ⁡ n → ∞ ∏ r = 1 n cos ⁡ 2 ( θ 2 r − 1 ) cos ⁡ 2 ( θ 2 r ) = lim ⁡ θ → 0 cos ⁡ 2 θ = 1 \begin{aligned} L&=\lim_{\theta\to 0}\lim_{n\to \infty}\prod_{r=1}^n\left [1 - \tan^2 \left(\frac\theta {2^r}\right )\right] \\&=\lim_{\theta\to 0}\lim_{n\to \infty}\prod_{r=1}^n \frac{\cos^2\left(\frac\theta {2^r}\right)-\sin^2\left(\frac\theta {2^r}\right)}{\cos^2\left(\frac\theta {2^r}\right)} \\&=\lim_{\theta\to 0}\lim_{n\to \infty}\prod_{r=1}^n \frac{\cos^2\left(\frac\theta {2^{r-1}}\right)}{\cos^2\left(\frac\theta {2^r}\right)} \\&=\lim_{\theta\to 0}\cos^2\theta \\&=\boxed 1 \end{aligned}

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check line 3!

Natalia Dcruz - 3 years ago

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